1) Use De Moivre’s theorem to show that $\displaystyle (1 + i \tan(x))^5 = \frac{1}{\cos^5(x)} cis (5x)$
2) Hence, find expressions for cos (5x) and sin (5x) in terms of tan(x) and cos(x).
1) Use De Moivre’s theorem to show that $\displaystyle (1 + i \tan(x))^5 = \frac{1}{\cos^5(x)} cis (5x)$
2) Hence, find expressions for cos (5x) and sin (5x) in terms of tan(x) and cos(x).
This should be:
$\displaystyle
(1 + i \tan(x))^5 = \frac{1}{\cos^5(x)} [\cos (5x) + i \sin(5x)]
$
As you have done part (1) we need only worry about part (2)2) Hence, find expressions for cos (5x) and sin (5x) in terms of tan(x) and cos(x).
$\displaystyle (1 + i \tan(x))^5 = \frac{1}{\cos^5(x)} (\cos (5x) + i \sin(5x))$
implies:
$\displaystyle \cos^5(x) (1 + i \tan(x))^5 = \cos (5x) + i \sin(5x)$
so expanding the second term on the left and equating real and imaginary
parts we will have the answer required - but I have to rush off now, may be
able to finish this later.
RonL
hi, CaptainBlack, thanks. For the actual equation i wrote "cis" which is not a typo, as you guessed "cis(x)" means "cos(x + i sin(x)". It is just an abbreviated form we commonly use. Maybe you don't use it where you are.
Now for the second part, is there a fairly simple way to expand (1 + i tan(x))^5 ? Expanding a degree 5 poly is a bit of a pain isn't it?
(1 + i tan(x))^5 = 1 + 5 (i tan(x)) + 10 (i tan(x))^2 + 10(i tan(x))^3 +
.........................5(i tan(x))^4 + (i tan(x))^5
.......................= 1 + 5 i tan(x) - 10 (tan(x))^2 - 10 i (tan(x))^3 +
.........................5(tan(x))^4 + i (tan(x))^5
.......................= [1 - 10 (tan(x))^2 + 5(tan(x))^4] +
..........................i [5 tan(x) - 10 (tan(x))^3 + (tan(x))^5]
RonL